This general equation you gave is a parabola that opens either up or down (U or ^ shaped, in your words) depending on the signs of $c$: if $c > 0$ it opens up, if $c < 0$ it opens down. You can get the formulas for the right/left ones (C and >) by switching $x$ and $y$, and again whether it opens right or left depending on whether $c$ is positive or negative, respectively. There many other sorts of parabolas too, though--for example, you might have a parabola at a 45-degree angle to the axes.
Bézier curves are a convenient way to produce parameterizations of parabolas: a quadratic Bézier is a (part of a) parabola. If $P_0$ and $P_2$ are points on the parabola and $P_1$ the intersection of the tangents at those points, the quadratic Bézier curve they define is given by $$\phi:t\mapsto(1-t)^2P_0+2t(1-t)P_1+t^2P_2.\tag{1}$$ (The parameter $t$ is usually taken to range from $0$ to $1$ for a Bézier patch.)
We can reproduce your parametrization by taking the vertex $P_0(0,0)$ and an end of the latus rectum $P_2(p,2p)$ as the points on the parabola. (Here I use the conventional name $p$ for this parameter instead of the $a$ in the question.) The tangent at the end of the latus rectum meets the parabola’s axis at a 45° angle, so our third control point will be $P_1(0,p)$. Plugging these into (1) we get $$(1-t)^2(0,0)+2t(1-t)(0,p)+t^2(p,2p)=(pt^2,2pt),$$ as required. As described here, parametrization of a parabola by a pair of quadratic polynomials has a nice symmetry about the vertex. Choosing the vertex as our first control point makes this symmetry quite simple.
To obtain the corresponding parameterization for a general parabola, you can either rotate and translate these three points to match the position and orientation of the given parabola, or compute them from other information that you have about the parabola. For example, if we have a parabola with vertex $P_0(x_0,y_0)$, focal length $p$ and axis direction $\theta$, we will have $P_1=P_0+(-p\sin\theta,p\cos\theta)$ and $P_2=P_0+(p\cos\theta-2p\sin\theta,2p\cos\theta+p\sin\theta)$, which gives the parameterization $$\begin{align}x&=x_0-2pt\sin\theta+pt^2\cos\theta \\ y&= y_0+2pt\cos\theta+pt^2\sin\theta.\end{align}$$
I’ll leave working out this parameterization for the general-form equation to you. As a hint, remember that for the parabola $y=ax^2+bx+c$, $p={1\over4a}$ and that a parabola’s vertex is halfway between its focus and directrix.
Best Answer
There is no "canonic" way to give a Cartesian system of equations for a parabola in 3D space. The simplest and oldest way is that of giving a parabola as intersection between a plane and a cone, see here for an example.
On the other hand, the locus of points whose distance from a given line (directrix) is the same as their distance from a given point (focus) is a parabolic cylinder, so you may find more natural to give the parabola as the intersection between this cylinder and the plane of focus and directrix.